How do you find the integral #int t^2(t^3+4)^(-1/2)dt# using substitution?

2 Answers
Mar 13, 2017

#2/3(t^3+4)^(1/2)+C# or #2/3sqrt(t^3+4)+C#

Explanation:

The key to solving any integral is to see what "type" of integral it could classify as. When I see an integral I try to ask myself if it is a substitution, integration by parts, trigonometric, trig sub, or partial fractions integral. In order to know whether or not to use the substitution method is whether or not the integral has both a function and its derivative present in the integral.

For this integral: #intt^2(t^3+4)^(-1/2)dt#, there is a #t^3# and a #t^2# which hints that we may want to incorporate #t^3# into our substitution since its derivative is present.

Let's do just that.
#intt^2/(t^3+4)^(1/2)dt#

Let #u=t^3+4#
so, #du=3t^2dt# or #1/3du=t^2dt#

#1/3int1/u^(1/2)du=1/3intu^(-1/2)du=(1/3)(2u^(1/2))+C=2/3u^(1/2)+C#

Putting #t# back in, we get:
#2/3(t^3+4)^(1/2)+C#

or #2/3sqrt(t^3+4)+C#

Mar 13, 2017

#intt^2(t^3+4)^(-1/2)dt=2/3(t^3+4)^(1/2)+C#

Explanation:

Because the function outside the bracket is a constant#xx#the derivative of the bracket , it suggests that we might be able to do this by 'inspection'.

#intt^2(t^3+4)^(-1/2)dt#

using the power rule for integration let us 'guess' that teh integral is of teh form

#y=(t^3+4)^(1/2)#

differentiate this using the chain rule:

#(dy)/(dx)=1/2xx3t^2(t^3+4)^(-1/2)#

comparing this with the original question we see that the only difference is the constant #3/2# so we adjust our 'guess' by a suitable value.

#:.intt^2(t^3+4)^(-1/2)dt=2/3(t^3+4)^(1/2)+C#

If one can spot integrals by inspection it can save a lot of work!.